Good morning. I was reading Tong's Quantum Field Theory course and got stuck on a somewhat stupid step. Essentially, considering the Lagrangian density $$ L = - F_{\mu \nu}F^{\mu \nu} + i \bar{\psi} \gamma^\mu (\partial_\mu +i A_\mu) \psi $$ in 1+1 dimension. This lagrangian is invariant under vectorial transformations ($e^{i \alpha}\psi$) and axial transformation ($\psi \rightarrow e^{e \alpha \gamma_5} \psi$). however, then he says that the associated conserved currents are given by $$ J_V^\mu = \bar{\psi} \gamma^\mu \psi, \quad J_A = \bar{\psi} \gamma^\mu \gamma^5 \psi $$ I tried to obtain they using Noether's theorem: $$ J^\mu = \frac{\partial L}{\partial (\partial_\mu\psi)} (\delta \psi) = (i \psi^*\gamma^0 \gamma^\mu)(i \alpha \psi) = - \alpha \bar{\psi} \gamma^\mu \psi $$ But this is not equal to the notes equation! I would greatly appreciate it if someone could give me some indication about the error I am making.
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$\begingroup$ What is happening in your equation after $\frac{\partial L}{\partial(\partial_\mu\psi)}(\delta\psi)$? Guide us through how you got each term and what happens at each equal sign. Your mistake is there. $\endgroup$– ɪdɪət strəʊləCommented Feb 21 at 9:25
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$\begingroup$ Of course, thank you! I did: $\frac{\partial }{\partial (\partial_\mu)} L (i \alpha) = \frac{\partial }{\partial (\partial_\mu)} (i \psi^* \gamma^0 \gamma^\mu \underbrace{\partial_\mu \psi}_{=1} + \overbrace{i \psi^* \gamma^0 \gamma^\mu A_\mu \psi}_{cancel}) (i \alpha) = - \psi^* \gamma^0 \gamma^\mu \psi$, the -1 comes from $i^2=-1$ $\endgroup$– GorgaCommented Feb 21 at 9:39
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1$\begingroup$ Usually the best way to obtain the conserved current is by variating the action and equalling your variation to 0. In any case, your final result is proportional to the desired J, so you should be able to just redefine $J' = \frac{J}{-\alpha} = \bar{\psi} \gamma^{\mu} \psi$. $\endgroup$– Gabriel Ybarra MarcaidaCommented Feb 21 at 10:16
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2$\begingroup$ @Gorga now this reads better (still a bit sloppy though). In your question you forgot the $\partial / \partial(\partial_\mu \psi)$. In the end you see you get the correct (vector) current, since, if $\partial_\mu J^\mu = 0$, $\partial_\mu (C J_\mu) = 0$, for any constant $C$. I.e. you can absorb the minus and the $\alpha$ as the previous commenter said. $\endgroup$– ɪdɪət strəʊləCommented Feb 21 at 10:37
1 Answer
Let me do the calculation just for the vectorial current to show you the derivation.
The fields transformations reads: \begin{eqnarray} \psi &\rightarrow& e^{i\alpha} \psi, \\ \bar{\psi} &\rightarrow& e^{-i\alpha} \bar{\psi}. \end{eqnarray} The Noether current is: $$J^\mu = \frac{\partial L}{\partial(\partial_\mu \psi)} \left.\frac{d(\delta\psi)}{d\alpha}\right|_{\alpha=0} + \frac{\partial L}{\partial(\partial_\mu \bar{\psi})} \left.\frac{d(\delta \bar{\psi})}{d\alpha}\right|_{\alpha=0}.$$
Where $\delta\psi = \psi' - \psi = \psi(e^{i\alpha} - 1)$ and $\delta \bar{\psi} = \bar{\psi}' - \bar{\psi} = \bar{\psi}(e^{-i\alpha} -1)$, then: \begin{eqnarray} \left.\frac{d(\delta\psi)}{d\alpha}\right|_{\alpha=0} &=& i\psi, \\ \left.\frac{d(\delta\psi^*)}{d\alpha}\right|_{\alpha=0} &=& -i\bar{\psi}. \end{eqnarray}
By other hand, the lagrangian derivatives are: \begin{eqnarray} \frac{\partial L}{\partial(\partial_\mu \psi)} &=& i\bar{\psi}\gamma^\mu, \\ \frac{\partial L}{\partial(\partial_\mu \bar{\psi})} &=& 0. \end{eqnarray}
Putting all together we obtain: $$J^\mu = -\bar{\psi}\gamma^\mu \psi.$$
T.
PS: I checked the course you are reading and saw that the convention of the transformations are $\psi \rightarrow e^{-i\alpha}\psi$, that provoques a change in the global sign and then the current is: $$J^\mu = \bar{\psi}\gamma^\mu \psi$$ Q.E.D.
